equation 1: (1+x)^k >1 +k*x + k*x^2
n=k+1 step we need to prove this equation:
equation 2: (1+x)^(k+1) >1 +(k+1)*x + (k+1)*x^2
Just multiply equation 1 by (1+x) which is > 0
(1+x)^(k+1) >(1 +k*x + k*x^2) (1+x)
Then develop the second member
n1 is the smallest whole number for which this inequality works :
(1+x)^n >1 +n*x + n*x^2
also i am given that x>0
find n1
and prove this inequality for every n=>n1 by induction.
the base case:
(1+x)^n1 >1+n1*x + n1*x^2
i think its correct because i was told that this inequality works for n1.
n=k step we presume that this equation is true :
equation 1: (1+x)^k >1 +k*x + k*x^2
n=k+1 step we need to prove this equation:
equation 2: (1+x)^(k+1) >1 +(k+1)*x + (k+1)*x^2
now i need to multiply equation1 by sum thing
and
do
if a<b<c
then a<c
how to do thing in this case?